Proof with Step-by-step explanation:
Since it is given that f(x) is decreasing thus we infer that
[tex]\frac{d}{dx}f(x)=f'(x)<0[/tex]
Similarly it is given that g(x) is decreasing thus we infer that
[tex]\frac{d}{dx}g(x)=g'(x)<0[/tex]
Now let [tex]y=fog=f[g(x)]\\\\[/tex]
Differentiating both sides with respect to 'x' and using chain rule of differentiation we get
[tex]dy/dx=\frac{d}{dx}f[g(x)]\\\\y'=f'[g((x)]\times g'(x)\\\\\\[/tex]
Now since both f'(x) and g'(x) are both negative(<0) we can infer that their product is always positive (>0)
[tex]\therefore y'(x)>0[/tex]
Thus fog(x) is an increasing function.
